The threshold frequency for certain metal is `v_0`. When light of frequency `2v_0` is incident on it, the maximum velocity of photoelectrons is `4xx10^(6) ms^(-1)`. If the frequency of incident radiation is increaed to `5v_0`, then the maximum velocity of photoelectrons will be

Options

A. `(4)/(5)xx10^(6)ms^(-1)`
B. `2xx10^(6)ms^(-1)`
C. `8xx10^(6)ms^(-1)`
D. `2xx10^(7)ms^(-1)`

Explanation

To solve the problem step by step, we will use the principles of the photoelectric effect and Einstein's photoelectric equation.

### Step 1: Understand the given information
- The threshold frequency for the metal is denoted as \( v_0 \).
- When light of frequency \( 2v_0 \) is incident on the metal, the maximum velocity of the emitted photoelectrons is \( v = 4 \times 10^6 \, \text{m/s} \).

### Step 2: Use Einstein's photoelectric equation
Einstein's photoelectric equation states that the maximum kinetic energy (KE) of the emitted photoelectrons is given by:
\[ KE_{max} = h\nu - h\nu_0 \]
Where:
- \( h \) is Planck's constant,
- \( \nu \) is the frequency of the incident light,
- \( \nu_0 \) is the threshold frequency.

### Step 3: Calculate the kinetic energy for the first case
For the first case where the frequency is \( 2v_0 \):
\[ KE_{max} = h(2v_0) - h(v_0) = h(2v_0 - v_0) = h v_0 \]
The kinetic energy can also be expressed in terms of the velocity of the photoelectrons:
\[ KE_{max} = \frac{1}{2} mv^2 \]
Setting these equal gives:
\[ \frac{1}{2} mv^2 = h v_0 \]
Substituting \( v = 4 \times 10^6 \, \text{m/s} \):
\[ \frac{1}{2} m (4 \times 10^6)^2 = h v_0 \]

### Step 4: Simplify the equation
Calculating \( (4 \times 10^6)^2 \):
\[ (4 \times 10^6)^2 = 16 \times 10^{12} \]
Thus, we have:
\[ \frac{1}{2} m (16 \times 10^{12}) = h v_0 \]
This simplifies to:
\[ 8 m \times 10^{12} = h v_0 \quad \text{(1)} \]

### Step 5: Calculate the kinetic energy for the second case
Now, we consider the second case where the frequency is \( 5v_0 \):
\[ KE_{max} = h(5v_0) - h(v_0) = h(5v_0 - v_0) = h(4v_0) \]
Using the kinetic energy expression:
\[ KE_{max} = \frac{1}{2} mv'^2 \]
Setting these equal gives:
\[ \frac{1}{2} mv'^2 = h(4v_0) \]

### Step 6: Substitute \( h v_0 \) from equation (1)
From equation (1), we know \( h v_0 = 8 m \times 10^{12} \). Therefore:
\[ h(4v_0) = 4 \times (8 m \times 10^{12}) = 32 m \times 10^{12} \]
Now substituting this into the kinetic energy equation:
\[ \frac{1}{2} mv'^2 = 32 m \times 10^{12} \]

### Step 7: Solve for \( v' \)
Dividing both sides by \( \frac{1}{2} m \):
\[ v'^2 = 64 \times 10^{12} \]
Taking the square root:
\[ v' = 8 \times 10^6 \, \text{m/s} \]

### Final Answer
The maximum velocity of the photoelectrons when the frequency of the incident radiation is increased to \( 5v_0 \) is:
\[ \boxed{8 \times 10^6 \, \text{m/s}} \]